LOOP ANALYSIS. determine all currents and Voltages in IT IS DUAL TO NODE ANALYSIS - IT FIRST DETERMINES ALL CURRENTS IN A CIRCUIT

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Ezra Cook
6 years ago
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1 LOOP ANALYSS The second systematic technique to determine all currents and oltages in a circuit T S DUAL TO NODE ANALYSS - T FRST DETERMNES ALL CURRENTS N A CRCUT AND THEN T USES OHM S LAW TO COMPUTE NECESSARY OLTAGES THERE ARE STUATON WHERE NODE ANALYSS S NOT AN EFFCENT TECHNQUE AND WHERE THE NUMBER OF EQUATONS REQURED BY THS NEW METHOD S SGNFCANTLY SMALLER

2 LOOPS, MESHES AND LOOP CURRENTS a b 3 c 7 f 6 e 5 d A BASC CRCUT EACH COMPONENT S CHARACTERZED BY TS OLTAGE ACROSS AND TS CURRENT THROUGH A LOOP S A CLOSED PATH THAT DOES NOT GO TWCE OER ANY NODE. THS CRCUT HAS THREE LOOPS fabef ebcde fabcdef A MESH S A LOOP THAT DOES NOT ENCLOSE ANY OTHER LOOP. fabef, ebcde ARE MESHES A LOOP CURRENT S A (FCTCOUS) CURRENT THAT S ASSUMED TO FLOW AROUND A LOOP,, 3 ARE LOOP CURRENTS A MESH CURRENT S A LOOP CURRENT ASSOCATED TO A MESH., ARE MESH CURRENTS N A CRCUT, THE CURRENT THROUGH ANY COMPONENT CAN BE EXPRESSED N TERMS OF THE LOOP CURRENTS NOT EERY LOOP CURRENT S REQURED TO COMPUTE ALL THE CURRENTS THROUGH COMPONENTS a b 3 7 c f 6 e 5 d A BASC CRCUT 3 FOR EERY CRCUT THERE S A MNMUM NUMBER OF LOOP CURRENTS THAT ARE NECESSARY TO COMPUTE EERY CURRENT N THE CRCUT. SUCH A COLLECTON S CALLED A MNMAL SET (OF LOOP CURRENTS).

3 FOR A GEN CRCUT LET B NUMBER OF BRANCHES N NUMBER OF NODES THE MNMUM REQURED NUMBER OF LOOP CURRENTS S L B ( N ) MESH CURRENTS ARE ALWAYS NDEPENDENT AN EXAMPLE B 7 N 6 L 7 (6 ) TWO LOOP CURRENTS ARE REQURED. THE CURRENTS SHOWN ARE MESH CURRENTS. HENCE THEY ARE NDEPENDENT AND FORM A MNMAL SET

4 WRTE THE MESH EQUATONS + v R BOOKKEEPNG BRANCHES 8 NODES 7 LOOP CURRENTS NEEDED i R AND WE ARE TOLD TO USE MESH CURRENTS! THS DEFNES THE LOOP CURRENTS TO BE USED + v v R3 DENTFY ALL OLTAGE DROPS ii R 3 R v + ( i i R + + v R ) v R5 i R i R 5 WRTE KL ON EACH MESH TOP MESH: v S + vr vs + vr BOTTOM: v + v v + v v 0 USE OHM S LAW R R5 R S3 R3 0

5 DEELOPNG A SHORTCUT WRTE THE MESH EQUATONS R R - R 3 WHENEER AN ELEMENT HAS MORE THAN ONE LOOP CURRENT FLOWNG THROUGH T WE COMPUTE NET CURRENT N THE DRECTON OF TRAEL R 5 R DRAW THE MESH CURRENTS. ORENTATON CAN BE ARBTRARY. BUT BY CONENTON THEY ARE DEFNED CLOCKWSE NOW WRTE KL FOR EACH MESH AND APPLY OHM S LAW TO EERY RESSTOR. AT EACH LOOP FOLLOW THE PASSE SGN CONENTON USNG LOOP CURRENT REFERENCE DRECTON + R + ( ) R + R R + R + ) R 0 3 (

6 LEARNNG EXAMPLE: FND o USNG LOOP ANALYSC SHORTCUT: POLARTES ARE NOT NEEDED. APPLY OHM S LAW TO EACH ELEMENT AS KL S BENG WRTTEN REARRANGE k 6k 6 k + 9 k 6 6 3* / k 0. 5mA k + 6k and add 5 ma EXPRESS ARABLE OF NTEREST AS FUNCTON OF LOOP CURRENTS O

7 LEARNNG EXTENSON. DRAW THE MESH CURRENTS. WRTE MESH EQUATONS MESH ( k + k + k ) k 3[ ] ( k + (k + 6k) (6 + 3 MESH ) DDE BY k. GET NUMBERS FOR COEFFCENTS ON THE LEFT AND ma ON THE RHS 3. SOLE EQUATONS 8 3[ ma] + 8 9[ ma] * / and add 30 33[ ma 33 ] O 6 k [ ] 5

8 CRCUTS WTH NDEPENDENT CURRENT SOURCES KL THERE S NO RELATONSHP BETWEEN AND THE SOURCE CURRENT! HOWEER... MESH CURRENT S CONSTRANED MESH EQUATON MESH ma BY NSPECTON k + 8 k k (ma) + 3 ma O 6k 8k 9 [ ] TO OBTAN APPLY KL TO ANY CLOSED PATH THAT NCLUDES 0.5

9 LEARNNG EXAMPLE COMPUTE O USNG MESH ANALYSS KL FOR o TWO MESH CURRENTS ARE DEFNED BY CURRENT SOURCES ma ma MESH 3 USE KL TO COMPUTE o 3 + k(ma) + k( ma) ma k 3

10 CURRENT SOURCES SHARED BY LOOPS - THE SUPERMESH APPROACH. WRTE CONSTRANT EQUATON DUE TO MESH CURRENTS SHARNG CURRENT SOURCES 3 ma 3. WRTE EQUATONS FOR THE OTHER MESHES ma. DEFNE A SUPERMESH BY (MENTALLY) REMONG THE SHARED CURRENT SOURCE. SELECT MESH CURRENTS 5. WRTE KL FOR THE SUPERMESH 6 + k3 + k + k( ) + k ( 3 ) 0 SUPERMESH NOW WE HAE THREE EQUATONS N THREE UNKNOWNS. THE MODEL S COMPLETE

11 DRAW MESH CURRENTS WRTE MESH EQUATONS. MESH MESH : k x + k + k( ) 0 x : + k + k( ) CONTROLLNG ARABLE N TERMS OF LOOP CURRENTS x 0 REPLACE AND REARRANGE 6k 6k 0 k + 6k SOLE FOR k 6mA O k [ ]

12 n the following we shall solve using loop analysis two circuits that had previously been solved using node analysis This is one circuit. we recap first the node analysis approach and then we solve using loop analysis

13 DETERMNE o SELECT LOOP CURRENTS USNG LOOP ANALYSS Write loop equations Loop : ma Loop 3: ma 3 Loop : + k + k( ) 0 X 3 Loop : + k( + ) + k 0 3 X Controlling variable: k ( ) X 3 k + k 6 ma, ma k 8 ariable of nterest O k 3 START SELECTON USNG MESHES SELECT A GENERAL LOOP TO AOD SHARNG A CURRENT SOURCE

14 LEARNNG EXAMPLE Find the current o RECAP OF NODE : 3 3 node: 6 (constraint eq.) X k k k k k 5 5 FND NODES AND SUPER NODES@ : X k k CONTROLLNG ARABLES X X k 7 eqs in 7 variables. ARABLE OF NTEREST O 5 k k

15 EXERCSE PROBLEM: MESH : ma MESH : ma 5 + k + k MESH : [ ] ( ) mA + ma ma 33 O 6k [ ] 5 MESH : k + k 0 6 ma 3 O 6k 8[ ]

RESISTIVE CIRCUITS MULTI NODE/LOOP CIRCUIT ANALYSIS

RESISTIVE CIRCUITS MULTI NODE/LOOP CIRCUIT ANALYSIS RESSTE CRCUTS MULT NODE/LOOP CRCUT ANALYSS DEFNNG THE REFERENCE NODE S TAL 4 THESTATEMENT 4 S MEANNGLES UNTL THE REFERENCE PONT S DEFNED BY CONENTON THE GROUND SYMBOL SPECFES THE REFERENCE PONT. ALL NODE